\(\int \frac {\sqrt {a+b \cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x))}{\sqrt {\cos (c+d x)}} \, dx\) [898]

Optimal result

Integrand size = 44, antiderivative size = 473 \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=-\frac {(a-b) \sqrt {a+b} (4 b B+a C) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 a b d}+\frac {\sqrt {a+b} (a C+2 b (2 B+C)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b d}-\frac {\sqrt {a+b} \left (4 a b B-a^2 C+4 b^2 C\right ) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{4 b^2 d}+\frac {(4 b B+a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{4 b d \sqrt {\cos (c+d x)}}+\frac {C \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{2 d} \] Output:

-1/4*(a-b)*(a+b)^(1/2)*(4*B*b+C*a)*cot(d*x+c)*EllipticE((a+b*cos(d*x+c))^( 
1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/ 
(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b))^(1/2)/a/b/d+1/4*(a+b)^(1/2)*(a*C+2*b 
*(2*B+C))*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+ 
c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d* 
x+c))/(a-b))^(1/2)/b/d-1/4*(a+b)^(1/2)*(4*B*a*b-C*a^2+4*C*b^2)*cot(d*x+c)* 
EllipticPi((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/cos(d*x+c)^(1/2),(a+b)/b,(-( 
a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+sec(d*x+c))/(a-b)) 
^(1/2)/b^2/d+1/4*(4*B*b+C*a)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b/d/cos(d*x 
+c)^(1/2)+1/2*C*cos(d*x+c)^(1/2)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/d
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 21.56 (sec) , antiderivative size = 1175, normalized size of antiderivative = 2.48 \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

Integrate[(Sqrt[a + b*Cos[c + d*x]]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/S 
qrt[Cos[c + d*x]],x]
 

Output:

(C*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + ((-4* 
a*(4*b*B + 3*a*C)*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + 
 b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c 
 + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])* 
Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a 
+ b)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]) - 4*a*(8*a*B + 4*b*C)*(( 
Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)]*Sqrt[-(((a + b)*Cos[c + d*x]*C 
sc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*C 
sc[c + d*x]*EllipticF[ArcSin[Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2 
)/a]/Sqrt[2]], (-2*a)/(-a + b)]*Sin[(c + d*x)/2]^4)/((a + b)*Sqrt[Cos[c + 
d*x]]*Sqrt[a + b*Cos[c + d*x]]) - (Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + 
 b)]*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos 
[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*EllipticPi[-(a/b), ArcSin[S 
qrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]/Sqrt[2]], (-2*a)/(-a + b) 
]*Sin[(c + d*x)/2]^4)/(b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]])) + 2 
*(4*b*B + a*C)*((I*Cos[(c + d*x)/2]*Sqrt[a + b*Cos[c + d*x]]*EllipticE[I*A 
rcSinh[Sin[(c + d*x)/2]/Sqrt[Cos[c + d*x]]], (-2*a)/(-a - b)]*Sec[c + d*x] 
)/(b*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[((a + b*Cos[c + d*x])*Sec[ 
c + d*x])/(a + b)]) + (2*a*((a*Sqrt[((a + b)*Cot[(c + d*x)/2]^2)/(-a + b)] 
*Sqrt[-(((a + b)*Cos[c + d*x]*Csc[(c + d*x)/2]^2)/a)]*Sqrt[((a + b*Cos[...
 

Rubi [A] (verified)

Time = 2.16 (sec) , antiderivative size = 471, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 3508, 3042, 3482, 3042, 3540, 25, 3042, 3532, 3042, 3288, 3477, 3042, 3295, 3473}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3508

\(\displaystyle \int \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)} (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3482

\(\displaystyle \frac {1}{4} \int \frac {(4 b B+a C) \cos ^2(c+d x)+2 (2 a B+b C) \cos (c+d x)+a C}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \int \frac {(4 b B+a C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 (2 a B+b C) \sin \left (c+d x+\frac {\pi }{2}\right )+a C}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3540

\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {-\left (\left (-C a^2+4 b B a+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b C \cos (c+d x)+a (4 b B+a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}+\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {-\left (\left (-C a^2+4 b B a+4 b^2 C\right ) \cos ^2(c+d x)\right )-2 a b C \cos (c+d x)+a (4 b B+a C)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {\left (C a^2-4 b B a-4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )+a (4 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3532

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B+a C)-2 a b C \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-\left (a^2 (-C)+4 a b B+4 b^2 C\right ) \int \frac {\sqrt {\cos (c+d x)}}{\sqrt {a+b \cos (c+d x)}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B+a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (a^2 (-C)+4 a b B+4 b^2 C\right ) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3288

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {a (4 b B+a C)-2 a b C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2 (-C)+4 a b B+4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3477

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (a C+4 b B) \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx-a (a C+2 b (2 B+C)) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx+\frac {2 \sqrt {a+b} \left (a^2 (-C)+4 a b B+4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {-a (a C+2 b (2 B+C)) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+a (a C+4 b B) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2 (-C)+4 a b B+4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3295

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {a (a C+4 b B) \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \left (a^2 (-C)+4 a b B+4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} (a C+2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {1}{4} \left (\frac {(a C+4 b B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {2 \sqrt {a+b} \left (a^2 (-C)+4 a b B+4 b^2 C\right ) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} (a C+2 b (2 B+C)) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{d}+\frac {2 (a-b) \sqrt {a+b} (a C+4 b B) \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right )}{a d}}{2 b}\right )+\frac {C \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}{2 d}\)

Input:

Int[(Sqrt[a + b*Cos[c + d*x]]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Sqrt[Co 
s[c + d*x]],x]
 

Output:

(C*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(2*d) + (-1/2 
*((2*(a - b)*Sqrt[a + b]*(4*b*B + a*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[ 
a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))] 
*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)] 
)/(a*d) - (2*Sqrt[a + b]*(a*C + 2*b*(2*B + C))*Cot[c + d*x]*EllipticF[ArcS 
in[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/( 
a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/ 
(a - b)])/d + (2*Sqrt[a + b]*(4*a*b*B - a^2*C + 4*b^2*C)*Cot[c + d*x]*Elli 
pticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c 
+ d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[( 
a*(1 + Sec[c + d*x]))/(a - b)])/(b*d))/b + ((4*b*B + a*C)*Sqrt[a + b*Cos[c 
 + d*x]]*Sin[c + d*x])/(b*d*Sqrt[Cos[c + d*x]]))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3288

Int[Sqrt[(b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[2*b*(Tan[e + f*x]/(d*f))*Rt[(c + d)/b, 2]*Sqrt[c 
*((1 + Csc[e + f*x])/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*Ellipti 
cPi[(c + d)/d, ArcSin[Sqrt[c + d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + 
 d)/b, 2]], -(c + d)/(c - d)], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - 
 d^2, 0] && PosQ[(c + d)/b]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

rule 3482

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[-2*B*Cos[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((c + d*Sin[e + f*x])^n/(f*(2* 
n + 3))), x] + Simp[1/(2*n + 3)   Int[((c + d*Sin[e + f*x])^(n - 1)/Sqrt[a 
+ b*Sin[e + f*x]])*Simp[a*A*c*(2*n + 3) + B*(b*c + 2*a*d*n) + (B*(a*c + b*d 
)*(2*n + 1) + A*(b*c + a*d)*(2*n + 3))*Sin[e + f*x] + (A*b*d*(2*n + 3) + B* 
(a*d + 2*b*c*n))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, 
B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Eq 
Q[n^2, 1/4]
 

rule 3508

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ 
.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*C, 0]
 

rule 3532

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(3/2)*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[C/b^2   Int[Sqrt[a + b*Sin[e + f*x]] 
/Sqrt[c + d*Sin[e + f*x]], x], x] + Simp[1/b^2   Int[(A*b^2 - a^2*C + b*(b* 
B - 2*a*C)*Sin[e + f*x])/((a + b*Sin[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x 
]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] & 
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3540

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(Sqrt[c + d*Sin[e + f 
*x]]/(d*f*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[1/(2*d)   Int[(1/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]))*Simp[2*a*A*d - C*(b*c - a*d) - 
 2*(a*c*C - d*(A*b + a*B))*Sin[e + f*x] + (2*b*B*d - C*(b*c + a*d))*Sin[e + 
 f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1183\) vs. \(2(422)=844\).

Time = 10.98 (sec) , antiderivative size = 1184, normalized size of antiderivative = 2.50

method	result	size
default	\(\text {Expression too large to display}\)	\(1184\)
parts	\(\text {Expression too large to display}\)	\(1220\)

Input:

int((a+b*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2), 
x,method=_RETURNVERBOSE)
 

Output:

-1/4/d*((8*cos(d*x+c)^2+16*cos(d*x+c)+8)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/ 
2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticPi(-csc(d*x 
+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(-2*cos(d*x+c)^2-4*cos(d*x+c)-2)*C 
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c) 
))^(1/2)*a^2*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,(-(a-b)/(a+b))^(1/2))+(8 
*cos(d*x+c)^2+16*cos(d*x+c)+8)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b 
)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^2*EllipticPi(-csc(d*x+c)+cot(d* 
x+c),-1,(-(a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8*cos(d*x+c)+4)*B*(cos(d*x+c 
)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a* 
b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(4*cos(d*x+c)^2+8 
*cos(d*x+c)+4)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c 
))/(1+cos(d*x+c)))^(1/2)*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b 
))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2 
)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*a^2*EllipticE(-csc(d*x+c 
)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(cos(d* 
x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos(d*x+c))/(1+cos(d*x+c)))^(1/2) 
*a*b*EllipticE(-csc(d*x+c)+cot(d*x+c),(-(a-b)/(a+b))^(1/2))+(-8*cos(d*x+c) 
^2-16*cos(d*x+c)-8)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(a+b*cos( 
d*x+c))/(1+cos(d*x+c)))^(1/2)*a*b*EllipticF(-csc(d*x+c)+cot(d*x+c),(-(a-b) 
/(a+b))^(1/2))+(2*cos(d*x+c)^2+4*cos(d*x+c)+2)*C*(cos(d*x+c)/(1+cos(d*x...
 

Fricas [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(1/2),x, algorithm="fricas")
 

Output:

integral((C*cos(d*x + c) + B)*sqrt(b*cos(d*x + c) + a)*sqrt(cos(d*x + c)), 
 x)
 

Sympy [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \left (B + C \cos {\left (c + d x \right )}\right ) \sqrt {a + b \cos {\left (c + d x \right )}} \sqrt {\cos {\left (c + d x \right )}}\, dx \] Input:

integrate((a+b*cos(d*x+c))**(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c 
)**(1/2),x)
 

Output:

Integral((B + C*cos(c + d*x))*sqrt(a + b*cos(c + d*x))*sqrt(cos(c + d*x)), 
 x)
 

Maxima [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(1/2),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)/sqr 
t(cos(d*x + c)), x)
 

Giac [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{\sqrt {\cos \left (d x + c\right )}} \,d x } \] Input:

integrate((a+b*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^ 
(1/2),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(b*cos(d*x + c) + a)/sqr 
t(cos(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}{\sqrt {\cos \left (c+d\,x\right )}} \,d x \] Input:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(1/2))/cos(c 
 + d*x)^(1/2),x)
 

Output:

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^(1/2))/cos(c 
 + d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b \cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\sqrt {\cos (c+d x)}} \, dx=\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right ) b +a}\, \sqrt {\cos \left (d x +c \right )}d x \right ) b \] Input:

int((a+b*cos(d*x+c))^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2), 
x)
 

Output:

int(sqrt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x))*cos(c + d*x),x)*c + int(sq 
rt(cos(c + d*x)*b + a)*sqrt(cos(c + d*x)),x)*b